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A polynomial given by the equation 4ax3 +3bx2+2cx+d=0 satisfy the condition 27a+9b+3c=0,then it has at least one real root lying in the interval
- a)(0,1)
- b)(3,4)
- c)(0,3)
- d)(4,5)
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A polynomial given by the equation 4ax3 +3bx2+2cx+d=0 satisfy the cond...
Explanation:
Given Polynomial:
The given polynomial is 4ax^3 + 3bx^2 + 2cx + d = 0.
Condition Given:
The condition given is 27a + 9b + 3c = 0.
Relation with Real Roots:
For a polynomial to have at least one real root in a given interval, the polynomial should change sign at the endpoints of the interval.
Sign Change Criterion:
The sign change criterion states that if a polynomial p(x) satisfies p(a) * p(b) < 0,="" then="" there="" is="" at="" least="" one="" real="" root="" between="" a="" and="" />
Application:
Substitute x = 1 and x = 0 into the given polynomial:
At x = 1, the polynomial becomes 4a + 3b + 2c + d = 0
At x = 0, the polynomial becomes d = 0
Using the Condition Given:
Substitute d = 0 into 4a + 3b + 2c + d = 0:
4a + 3b + 2c = 0
Dividing by 3, we get: 27a + 9b + 3c = 0
Sign Change Criterion Check:
Since 27a + 9b + 3c = 0, the polynomial changes sign between x = 0 and x = 1.
Hence, the polynomial has at least one real root lying in the interval (0, 1).
Therefore, the correct answer is option (a) (0, 1).
Given Polynomial:
The given polynomial is 4ax^3 + 3bx^2 + 2cx + d = 0.
Condition Given:
The condition given is 27a + 9b + 3c = 0.
Relation with Real Roots:
For a polynomial to have at least one real root in a given interval, the polynomial should change sign at the endpoints of the interval.
Sign Change Criterion:
The sign change criterion states that if a polynomial p(x) satisfies p(a) * p(b) < 0,="" then="" there="" is="" at="" least="" one="" real="" root="" between="" a="" and="" />
Application:
Substitute x = 1 and x = 0 into the given polynomial:
At x = 1, the polynomial becomes 4a + 3b + 2c + d = 0
At x = 0, the polynomial becomes d = 0
Using the Condition Given:
Substitute d = 0 into 4a + 3b + 2c + d = 0:
4a + 3b + 2c = 0
Dividing by 3, we get: 27a + 9b + 3c = 0
Sign Change Criterion Check:
Since 27a + 9b + 3c = 0, the polynomial changes sign between x = 0 and x = 1.
Hence, the polynomial has at least one real root lying in the interval (0, 1).
Therefore, the correct answer is option (a) (0, 1).
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Community Answer
A polynomial given by the equation 4ax3 +3bx2+2cx+d=0 satisfy the cond...
f(x)=ax4+ bx3+ cx2+dx+e
∵ f(0)=f(3)=e
∴ f′(x) become vanishes in (0,3)
∵ f(0)=f(3)=e
∴ f′(x) become vanishes in (0,3)
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A polynomial given by the equation 4ax3 +3bx2+2cx+d=0 satisfy the condition 27a+9b+3c=0,then it has at least one real root lying in the intervala)(0,1)b)(3,4)c)(0,3)d)(4,5)Correct answer is option 'C'. Can you explain this answer?
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